Photons collected be telescope apertures

[trina1990]

the energy output of a star is such at the surface of the Earth it provides 5x10^-19 WM^-2 in the visible part of the spectrum...
how many photons are collected per second by a 500mm diameter telescope?

i don't know if there's any definite formula to drive the answer...but i tried this way..
stellar flux=5x10^-19WM^-2=5x10^-25 J/S/mm^2
E=hc/lambda joul/photon ( assuming lambda to be 550nm)

then flux/E=1.383x10^-6 photon /S/mm^2

then i found the telescope mirror are to be 4pi (500)^2=3.142x10^6 mm^2
now i derived the number of photons per second collected by the 500mm diameter to be 4.4!

can this be right?
is there any other way to solve this?

 

[cepheid]

Aside from the fact that the area of a circle is $\pi r^2$ (without the factor of 4), I don't see any problems with your math.

I did some rough calculations, and found that a star with that flux has an apparent magnitude of something like +26. Again, I want to emphasize that this was a very rough calculation. I computed the flux of Vega using its stated luminosity and distance (using Wikipedia numbers), and then took m = -2.5log(F_star / F_vega) to estimate the magnitude. I know that Vega isn't used as the zero point for V band magnitudes anymore, but this was just a back of the envelope calculation to show that you're considering a very dim star. In fact Wikipedia says that magnitude 27 is at the limit of what 8-metre ground-based telescopes can see, so it seems plausible that such a star would be invisible in a 0.5-metre telescope (esp. if your detector is the human eye).

 

[trina1990]

thanks a lot...
the book in which i found this problem showed answer of this problem to be 3x10^7 photons per second...
that's why i got so much stunned with my result...

but a similar problem with different coefficients shows a result very close to my calculation..
anyway i understood the err of my measuring the area of the aperture...
& with the information you provided from wikipedia, then it's most likely to support the answer...
thanks again